Question

A catfish is 4.73 m below the surface of a smooth lake.

(a) What is the diameter of the circle on the surface through which the fish can see the world outside the water?
_____m

Answer 1

The equation for light bending at the surface between two materials is Snell's law: n1/n2 = sin(theta2)/sin(theta1), where theta is the angle at which the light hits the surface, and n is the index of refraction. Lets say 1 refers to air and 2 refers to water. We take n1=1 (since it's almost 1 in air), and n2=1.329, as given by the problem.

Light coming to/from the fish will bend outward when it goes from water to air. If the angle in the water (theta2) is big enough, then it will bend out so far in the air that it won't be going through the air at all, but reflecting back into the water. That would happen when theta1=90 degrees. So what theta1 gives us a theta1 of 90? Just plug into the equation:

1/1.329 = sin(theta2)/sin(90)
0.7524=sin(theta2)
theta2=48.80 degrees

So the fish can see everything inside of a cone of angle 48.80 degrees, and a depth from the surface of 4.73m. The radius of that cone is just trigonometry: tan(theta2)=r/4.73, so r=5.403 m. The diameter is twice the radius, or 10.806m.